As the title say, I want to prove that $C_c^{\infty}(\mathbb R^n)$ is dense in $W^{k,p}(\mathbb R^n)$
i.e. $\displaystyle{ W^{k,p} (\mathbb R^n) = W_0^{k,p}(\mathbb R^n) \quad (\star)}$.
In a book I found that in order to prove it, we need the following claim:
Claim: Let $\displaystyle{\zeta \in C^{\infty} ([0, \infty)) }$ such that $\zeta (t) =1 $ for $ t \leq 1 $ and $\displaystyle{ \zeta (t) =0 }$ for $ t \geq 2$. For $R>0$ and $x \in \mathbb R^n$ define $\displaystyle{ \eta_R (x) = \zeta \left( \frac{|x|}{R} \right)}$. If $ \displaystyle{ u \in W^{k,p} ( \mathbb R^n) }$ with $1 \leq p < \infty $ then, $\displaystyle{ \eta_R u \to u }$ in $W^{k,p}(\mathbb R^n) $ as $ R \to \infty$.
If I prove the claim then the $(\star)$ follows using mollifiers. But unfortuently I have stuck, with the proof of the claim.
Any ideas?
It may help to consider first the case $n=1$ and $k=2$. The fact that $\eta_Ru\to u$ in $L^p$ follows from a monotone convergence argument. Indeed, since $\eta_R(x)=1$ if $|x|\leqslant R$, $$\int_\mathbb R|\eta_R(x)u(x)-u(x)|^p\mathrm dx=\int_{\{|x|\gt R\}}|u(x)|^p\mathrm dx.$$
We can use the product rule for derivatives: we get $(\eta_R u)'=\eta_R u'+\eta'_Ru$. The first terms goes to $u'$ by a similar argument as before. For the second one, there is a term $\frac 1R$ which appears when we take the derivative. Since the derivative of $\eta$ is bounded, the light version of the claim follows.
For the general case, we have to consider Leibnitz rule. It's the same idea, altought it's more technical. We shall need that each derivative of $\eta$ is bounded.